1. Nov 22, 2009

### qwert13

1. The problem statement, all variables and given/known data
The main water line enters a house on the first floor. The line has a gauge pressure of 1.90 multiplied by 105 Pa.
(a) A faucet on the second floor, 6.3 m above the first floor, is turned off. What is the gauge pressure at this faucet?
(b) How high could a faucet be before no water would flow from it, even if the faucet were open?

2. Relevant equations
P=F/A
density= m/v
Fbuoyancy= (density)Vg
P + (1/2)(density)(v)^2 + (density)gh = constant
P= (density)gh

3. The attempt at a solution
P= 1000(9.8)(6.3) = 61740? not right. i'm not sure what else to do.

2. Nov 24, 2009

### qwert13

Anyone? It's due tonight so if someone could give me a couple of tips that would be great. thanks in advance

3. Nov 24, 2009

### ideasrule

1. (a) It's 61740 Pa below the gauge pressure on the lower floor, since the pressure applied by the 6.3 m water column is part of the water pressure felt on the first floor. Are you sure the line's gauge pressure is 1.90*105 Pa and not 1.90*105 kPa?

(b) What's the minimum water pressure necessary to force water out?

4. Nov 24, 2009

### qwert13

Yeah actually the pressure is 1.9 x10^5 Pa not 1.9 x 105 Pa. I accidentally typed it wrong onto the computer

5. Nov 24, 2009

### qwert13

ok I figured it out. Thank you so much ideasrule for your help.